Right quadratic variation for stochastic process Say we have a filtered prob. space, $M$ an $L^2$ martingale on it and $Y_{a_i}$ $\mathcal{F}_{a_i}$ measurable random variable. Why does it hold that
$$E[(Y_{a_i}(M_{t \wedge a_{i+1}}-M_{t \wedge a_i}))^2] = E[Y_{a_i}^2(\langle M \rangle_{t \wedge a_{i+1}}- \langle M \rangle_{t\wedge a_i})]$$
The measurability of $Y_{a_i}$ is clear but does the equality work because the difference of quadratic variations on the right-hand side is the quadratic variation of $(M_{t \wedge a_{i+1}}-M_{t \wedge a_i})$? If so, why is that?
 A: You say you understand the orthogonality of $Y_{a_t}$ (not necessarily independence), so I will show why $E[(M_{t \wedge a_{t+1}}-M_{t \wedge a_t})^2] = E[\langle M \rangle_{t \wedge a_{t+1}} - \langle M \rangle_{t \wedge a_t}]$.  By definition, $M_t^2 - \langle M\rangle_t$ is a martingale, and since a stopped martingale is a martingale this implies $M_{t \wedge a_{t+1}}^2 - \langle M \rangle_{t \wedge a_{t+1}}$ and $M_{t \wedge a_t}^2-\langle M \rangle_{t \wedge a_t}$ are as well.  Thus we compute
\begin{align*}
E[(M_{t \wedge a_{t+1}}-M_{t \wedge a_t})^2] &= E[M_{t \wedge a_{t+1}}^2-2 M_{t \wedge a_{t+1}}M_{t \wedge a_t} + M_{t \wedge a_{t}}^2] \\
&= E[M_{t \wedge a_{t+1}}^2 - \langle M \rangle_{t \wedge a_{t+1}} + \langle M \rangle_{t \wedge a_{t+1}} -2 M_{t \wedge a_{t+1}}M_{t \wedge a_t} + M_{t \wedge a_{t}}^2] \\
&= E[E[M_{t \wedge a_{t+1}}^2 - \langle M \rangle_{t \wedge a_{t+1}} | \mathcal F_{a_t}] + \langle M \rangle_{t \wedge a_{t+1}} -E[2 M_{t \wedge a_{t+1}}M_{t \wedge a_t}|\mathcal F_{a_t}] + M_{t \wedge a_{t}}^2] \\
&= E[M_{t \wedge a_t}^2 - \langle M \rangle_{t \wedge a_{t}}+ \langle M \rangle_{t \wedge a_{t+1}} -2 M_{t \wedge a_t}E[ M_{t \wedge a_{t+1}}|\mathcal F_{a_t}] + M_{t \wedge a_{t}}^2] \\
&= E[2 M_{t \wedge a_t}^2 + \langle M \rangle_{t \wedge a_{t+1}} - \langle M \rangle_{t \wedge a_{t}} - 2 M_{t \wedge a_t}^2] \\
&= E[\langle M \rangle_{t \wedge a_{t+1}} - \langle M \rangle_{t \wedge a_{t}}].
\end{align*}
A similar computation would show that $(M_{t \wedge a_{t+1}}-M_{t \wedge a_t})^2 - (\langle M \rangle_{t \wedge a_{t+1}} - \langle M \rangle_{t \wedge a_{t}})$ is a martingale, and $\langle M \rangle_{t \wedge a_{t+1}} - \langle M \rangle_{t \wedge a_{t}}$ is an increasing process, so the difference of the quadratic variations is indeed the quadratic variation of $M_{t \wedge a_{t+1}}-M_{t \wedge a_t}$.
